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Leila is playing a carnival game in which she is given 4 chances to
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04 May 2021, 03:36
04 May 2021, 03:36
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38% (02:19) wrong based on 18 sessions
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Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A. \(\frac{1}{5^4}\)
B. \(\frac{1}{5^3}\)
C. \(\frac{6}{5^4}\)
D. \(\frac{13}{5^4}\)
E. \(\frac{17}{5^4}\)
A. \(\frac{1}{5^4}\)
B. \(\frac{1}{5^3}\)
C. \(\frac{6}{5^4}\)
D. \(\frac{13}{5^4}\)
E. \(\frac{17}{5^4}\)
Re: Leila is playing a carnival game in which she is given 4 chances to
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04 May 2021, 06:20
04 May 2021, 06:20
1
Probability of success on atleast 3 throws = 4C3 * (1/5)^3*4/5 + (1/5)^4 = 17/5^4
Answer is E
Answer is E
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Leila is playing a carnival game in which she is given 4 chances to
[#permalink]
04 May 2021, 08:45
04 May 2021, 08:45
1
GeminiHeat wrote:
Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A. \(\frac{1}{5^4}\)
B. \(\frac{1}{5^3}\)
C. \(\frac{6}{5^4}\)
D. \(\frac{13}{5^4}\)
E. \(\frac{17}{5^4}\)
A. \(\frac{1}{5^4}\)
B. \(\frac{1}{5^3}\)
C. \(\frac{6}{5^4}\)
D. \(\frac{13}{5^4}\)
E. \(\frac{17}{5^4}\)
P(win) = \(\frac{1}{5}\)
P(lose) = \(\frac{4}{5}\)
win on all 4 throws = P(4 wins)= \((\frac{1}{5})(\frac{1}{5})(\frac{1}{5})(\frac{1}{5}) = \frac{1}{(5^4)}\)
win on 3 throws = P(3 wins) P(1 lose) = \((\frac{1}{5})(\frac{1}{5})(\frac{1}{5})(\frac{4}{5}) = \frac{4}{(5^4)}\)
Note: win on 3 throws can be arranged in 4 ways: WWWL, WWLW, WLWW, LWWW
So, Final Prob. = \(4(\frac{4}{5^4}) + \frac{1}{(5^4)} = \frac{17}{(5^4)}\)
Hence, option E
